Pade Approximation Proof

242 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION

Pade Approximations

4.6 Pade Approximations

In this section we introduce the notion of rational approximations for functions. Thefunction f (x) will be approximated over a small portion of its domain. For example,if f (x) = cos(x), it is sufficient to have a formula to generate approximations on theinterval [0, π/2]. Then trigonometric identities can be used to compute cos(x) for anyvalue x that lies outside [0, π/2].

SEC. 4.6 PADE APPROXIMATIONS 243

A rational approximation to f (x) on [a, b] is the quotient of two polynomialsPN (x) and QM (x) of degrees N and M , respectively. We use the notation RN ,M (x) todenote this quotient:

(1) RN ,M (x) = PN (x)

QM (x)for a ≤ x ≤ b.

Our goal is to make the maximum error as small as possible. For a given amountof computational effort, one can usually construct a rational approximation that has asmaller overall error on [a, b] than a polynomial approximation. Our development isan introduction and will be limited to Pade approximations.

The method of Pade requires that f (x) and its derivative be continuous at x = 0.There are two reasons for the arbitrary choice of x = 0. First, it makes the manipula-tions simpler. Second, a change of variable can be used to shift the calculations over toan interval that contains zero. The polynomials used in (1) are

(2) PN (x) = p0 + p1x + p2x2 + · · · + pN x N

and

(3) QM (x) = 1+ q1x + q2x2 + · · · + qM x M .

The polynomials in (2) and (3) are constructed so that f (x) and RN ,M (x) agree atx = 0 and their derivatives up to N + M agree at x = 0. In the case Q0(x) = 1, theapproximation is just the Maclaurin expansion for f (x). For a fixed value of N + Mthe error is smallest when PN (x) and QM (x) have the same degree or when PN (x) hasdegree one higher than QM (x).

Notice that the constant coefficient of QM is q0 = 1. This is permissible, becauseit cannot be 0 and RN ,M (x) is not changed when both PN (x) and QM (x) are dividedby the same constant. Hence the rational function RN ,M (x) has N + M + 1 unknowncoefficients. Assume that f (x) is analytic and has the Maclaurin expansion

(4) f (x) = a0 + a1x + a2x2 + · · · + ak xk + · · · ,

and form the difference f (x)QM (x)− PN (x) = Z(x):

(5)

∞∑j=0

a j x j

M∑j=0

q j x j

− N∑j=0

p j x j =∞∑

j=N+M+1

c j x j .

The lower index j = M + N + 1 in the summation on the right side of (5) is chosenbecause the first N + M derivatives of f (x) and RN ,M (x) are to agree at x = 0.

When the left side of (5) is multiplied out and the coefficients of the powers of x j

are set equal to zero for k = 0, 1, . . . , N + M , the result is a system of N + M + 1


y = R4,4(x)

y = cos(x)

−1 1 2 3 4 5−2−3−4−5x

0.5

1.0

−0.5

−1.0

y

Figure 4.18 The graph of y = cos(x) and its Padeapproximation R4,4(x).

linear equations:

a0 − p0 = 0

q1a0 + a1 − p1 = 0

q2a0 + q1a1 + a2 − p2 = 0(6)

q3a0 + q2a1 + q1a2 + a3 − p3 = 0

qM aN−M + qM−1aN−M+1 + · · · + aN − pN = 0

and

(7)

qM aN−M+1 + qM−1aN−M+2 + · · · + q1aN + aN+1 = 0

qM aN−M+2 + qM−1aN−M+3 + · · · + q1aN+1 + aN+2 = 0

......

qM aN + qM−1aN+1 + · · · + q1aN+M−1 + aN+M = 0.

Notice that in each equation the sum of the subscripts on the factors of each productis the same, and this sum increases consecutively from 0 to N + M . The M equationsin (7) involve only the unknowns q1, q2, . . . , qM and must be solved first. Then theequations in (6) are used successively to find p0, p1, . . . , pN .

Example 4.17. Establish the Pade approximation

(8) cos(x) ≈ R4,4(x) = 15,120− 6900x2 + 313x4

15,120+ 660x2 + 13x4.

See Figure 4.18 for the graphs of cos(x) and R4,4(x) over [−5, 5].


If the Maclaurin expansion for cos(x) is used, we will obtain nine equations in nineunknowns. Instead, notice that both cos(x) and R4,4(x) are even functions and involvepowers of x2. We can simplify the computations if we start with f (x) = cos(x1/2):

(9) f (x) = 1− 1

2x + 1

24x2 − 1

720x3 + 1

40,320x4 − · · · .

In this case, equation (5) becomes(1− 1

2x + 1

24x2 − 1

720x3 + 1

40,320x4 − · · ·

)(1+ q1x + q2x2

)− p0− p1x− p2x2

= 0+ 0x + 0x2 + 0x3 + 0x4 + c5x5 + c6x6 + · · · .

When the coefficients of the first five powers of x are compared, we get the followingsystem of linear equations:

1− p0 = 0

−1

2+ q1 − p1 = 0

1

24− 1

2q1 + q2 − p2 = 0(10)

− 1

720+ 1

24q1 − 1

2q2 = 0

1

40,320− 1

720q1 + 1

24q2 = 0.

The last two equations in (10) must be solved first. They can be rewritten in a form that iseasy to solve:

q1 − 12q2 = 1

30and − q1 + 30q2 = −1

56.

First find q2 by adding the equations; then find q1:

(11)q2 = 1

18

(1

30− 1

56

)= 13

15,120,

q1 = 1

30+ 156

15,120= 11

252.

Now the first three equations of (10) are used. It is obvious that p0 = 1, and we canuse q1 and q2 in (11) to solve for p1 and p2:

(12)p1 = −1

2+ 11

252= −115

252,

p2 = 1

24− 11

504+ 13

15,120= 313

15,120.

Now use the coefficients in (11) and (12) to form the rational approximation to f (x):

(13) f (x) ≈ 1− 115x/252+ 313x2/15,120

1+ 11x/252+ 13x2/15,120.


Since cos(x) = f (x2), we can substitute x2 for x in equation (13) and the result is theformula for R4,4(x) in (8). �

Continued Fraction Form

The Pade approximation R4,4(x) in Example 4.17 requires a minimum of 12 arithmeticoperations to perform an evaluation. It is possible to reduce this number to seven bythe use of continued fractions. This is accomplished by starting with (8) and findingthe quotient and its polynomial remainder.

R4,4(x) = 15,120/313− (6900/313)x2 + x4

15,120/13+ (660/13)x2 + x4

= 313

13−(

296,280

169

)(12,600/823+ x2

15,120/13+ (600/13)x2 + x4

).

The process is carried out once more using the term in the previous remainder. Theresult is

R4,4(x) = 313

13− 296,280/169

15,120/13+ (660/13)x2 + x4

12,600/823+ x2

= 313

13− 296,280/169

379,380

10,699+ x2 + 420,078,960/677,329

12,600/823+ x2

.

The fractions are converted to decimal form for computational purposes and we obtain

(14) R4,4(x) = 24.07692308

− 1753.13609467

35.45938873+ x2 + 620.19928277/(15.30984204+ x2).

To evaluate (14), first compute and store x2, then proceed from the bottom right termin the denominator and tally the operations: addition, division, addition, addition, divi-sion, and subtraction. Hence it takes a total of seven arithmetic operations to evaluateR4,4(x) in continued fraction form in (14).

We can compare R4,4(x) with the Taylor polynomial P6(x) of degree N = 6,which requires seven arithmetic operations to evaluate when it is written in the nestedform

(15)P6(x) = 1+ x2

(−1

2+ x2

(1

24− 1

720x2))

= 1+ x2(−0.5+ x2(0.0416666667− 0.0013888889x2)).


y = ER(x)

−0.5 0.0

(a)

0.5 1.0−1.0x

−1 × 10 −7

−2 × 10 −7

−3 × 10 −7

y = EP(x)

(b)

−0.5 0.0 0.5 1.0−1.0x

0.000005

0.000010

0.000015

0.000020

0.000025

Figure 4.19 (a) The graph of the error ER(x) = cos(x) − R4,4(x) for the Pade ap-proximation R4,4(x). (b) The graph of the error EP (x) = cos(x)− P6(x) for the Taylorapproximation P6(x).

The graphs of ER(x) = cos(x) − R4,4(x) and EP(x) = cos(x) − P6(x) over [−1, 1]are shown in Figure 4.19(a) and (b), respectively. The largest errors occur at theendpoints and are ER(1) = −0.0000003599 and EP(1) = 0.0000245281, respec-tively. The magnitude of the largest error for R4,4(x) is about 1.467% of the errorfor P6(x). The Pade approximation outperforms the Taylor approximation better onsmaller intervals, and over [−0.1, 0.1] we find that ER(0.1) = −0.0000000004 andEP(0.1) = 0.0000000966, so the magnitude of the error for R4,4(x) is about 0.384%of the magnitude of the error for P6(x).

Numerical Methods Using Matlab, 4th Edition, 2004 John H. Mathews and Kurtis K. Fink

ISBN: 0-13-065248-2

Prentice-Hall Inc. Upper Saddle River, New Jersey, USA

http://vig.prenhall.com/

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Pade Approximation Proof